Sometimes a very complex set of gates can be simplified to save on cost and make faster circuits. A quick way to do that is through boolean identities. Boolean identities are quick rules that allow you to simplify boolean expressions. For all situations described below:

A = It is raining upon the British Museum right now (or any other statement that can be true or false)
B = I have a cold (or any other statement that can be true or false)

Identity

Explanation

Truth Table

A.A=A{\displaystyle A.A=A}

It is raining AND It is raining is the same as saying It is raining

A{\displaystyle A}

A{\displaystyle A}

A.A{\displaystyle A.A}

0

0

0

1

1

1

A.A¯=0{\displaystyle A.{\overline {A}}=0}

It is raining AND It isn't raining is impossible at the same time so the statement is always false

A{\displaystyle A}

A¯{\displaystyle {\overline {A}}}

A.A¯{\displaystyle A.{\overline {A}}}

0

1

0

1

0

0

1+A=1{\displaystyle 1+A=1}

2+2=4 OR It is raining. So it doesn't matter whether it's raining or not as 2+2=4 and it is impossible to make the equation false

1

A{\displaystyle A}

1+A{\displaystyle 1+A}

1

0

1

1

1

1

0+A=A{\displaystyle 0+A=A}

1+2=4 OR It is raining. So it doesn't matter about the 1+2=4 statement, the only thing that will make the statement true or not is whether it's raining

0{\displaystyle 0}

A{\displaystyle A}

0+A{\displaystyle 0+A}

0

0

0

0

1

1

A+A=A{\displaystyle A+A=A}

It is raining OR It is raining is the equivalent of saying It is raining

A{\displaystyle A}

A{\displaystyle A}

A+A{\displaystyle A+A}

0

0

0

1

1

1

A+A¯=1{\displaystyle A+{\overline {A}}=1}

It is raining OR It isn't raining is always true

A{\displaystyle A}

A¯{\displaystyle {\overline {A}}}

A+A¯{\displaystyle A+{\overline {A}}}

0

1

1

1

0

1

0.A=0{\displaystyle 0.A=0}

1+2=4 AND It is raining. It is impossible to make 1+2=4 so this equation so this equation is always false

0{\displaystyle 0}

A{\displaystyle A}

0.A{\displaystyle 0.A}

0

0

0

0

1

0

1.A=A{\displaystyle 1.A=A}

2+2=4 AND It is raining. This statement relies totally on whether it is raining or not, so we can ignore the 2+2=4 part

1{\displaystyle 1}

A{\displaystyle A}

1.A{\displaystyle 1.A}

1

0

0

1

1

1

A+B=B+A{\displaystyle A+B=B+A}

It is raining OR I have a cold, is the same as saying: I have a cold OR It is raining

A{\displaystyle A}

B{\displaystyle B}

A+B{\displaystyle A+B}

B+A{\displaystyle B+A}

0

0

0

0

0

1

1

1

1

0

1

1

1

1

1

1

A.B=B.A{\displaystyle A.B=B.A}

It is raining AND I have a cold, is the same as saying: I have a cold AND It is raining

A{\displaystyle A}

B{\displaystyle B}

A.B{\displaystyle A.B}

B.A{\displaystyle B.A}

0

0

0

0

0

1

0

0

1

0

0

0

1

1

1

1

A+(A.B)=A{\displaystyle A+(A.B)=A}

It is raining OR (It is raining AND I have a cold). If It is raining then both sides of the equation are true. Or if It is not raining then both sides are false. Therefore everything relies on A and we can replace the whole thing with A. Alternatively we could play with the boolean algebra equation:

A+(A.B)=(1.A)+(A.B){\displaystyle A+(A.B)=(1.A)+(A.B)} Using the identity rule 1.A=A{\displaystyle 1.A=A}(1.A)+(A.B)=A.(1+B){\displaystyle (1.A)+(A.B)=A.(1+B)} Take out the A, common to both sides of the equationA.(1+B)=A.1{\displaystyle A.(1+B)=A.1} Using the identity rule 1+B=1{\displaystyle 1+B=1}A.1=A{\displaystyle A.1=A} Using the identity rule 1.A=A{\displaystyle 1.A=A}

A{\displaystyle A}

B{\displaystyle B}

A.B{\displaystyle A.B}

A+(A.B){\displaystyle A+(A.B)}

0

0

0

0

0

1

0

0

1

0

0

1

1

1

1

1

A.(A+B)=A{\displaystyle A.(A+B)=A}

It is raining AND (It is raining OR I have a cold). If It is raining then both sides of the equation are true. Or if It is not raining then both sides are false. Therefore everything relies on A and we can replace the whole thing with A. Alternatively we could play with the boolean algebra equation:

A.(A+B)=(0+A).(A+B){\displaystyle A.(A+B)=(0+A).(A+B)} Using the identity rule 0+A=A{\displaystyle 0+A=A}(0+A).(A+B)=A+(0.B){\displaystyle (0+A).(A+B)=A+(0.B)} Take out the A, common to both sides of the equationA+(0.B)=A+0{\displaystyle A+(0.B)=A+0} Using the identity rule 0.B=0{\displaystyle 0.B=0}A+0=A{\displaystyle A+0=A} Using the identity rule 0+A=A{\displaystyle 0+A=A}

B.(A+(A.B)){\displaystyle B.(A+(A.B))} treat the brackets first and the AND inside the brackets first
(B.A)+(B.A.B){\displaystyle (B.A)+(B.A.B)} multiply it out
(B.A)+(A.B){\displaystyle (B.A)+(A.B)} as B.A.B=A.B{\displaystyle B.A.B=A.B}A.B{\displaystyle A.B} as (B.A)=(A.B){\displaystyle (B.A)=(A.B)}

(A+B).(A+A){\displaystyle (A+B).(A+A)}

Answer:

(A+B).(A+A){\displaystyle (A+B).(A+A)}

(A+B).A{\displaystyle (A+B).A} as A=A+A{\displaystyle A=A+A}(A+B).(A+0){\displaystyle (A+B).(A+0)} as A=A+0{\displaystyle A=A+0}A+(B.0){\displaystyle A+(B.0)} take A out as the common denominator
A{\displaystyle A} as (B.0)=0{\displaystyle (B.0)=0}